Riddle

[the_nick]

a pound is a pound , but I prefer more the pound of gold

[crowebird]

lol Same here

[kingfisher-blue]

Each genuine coin weighs 1 gram, each fake coin weighs 1.5 grams no two bags have the same amount of coins in them, so you can`t
use balance scales.

You simply take 1 coin from bag one, two coins from bag 2 and three coins from bag 3. Now weigh the 6 coins. The answer lies in the
weight of the 6 coins you would be weighing.
If the fake coin was in bag 1 then the total weight would be 6.5 grams
If the fake coin was in bag 2 then the total weight would be 7 grams
if the fake coin was in bag three the total weight would then be 7.5 grams

[fruitfly]

Hi there again!
Seems like you like asking the riddles more than answering them??? I know I do!
But often the answer can be more brainteasing than the question! So here's the answer to the avgas/euro super 95 mixture:
Imagine that the wine and water separate out. It doesn't need to in reality, but we force it to in our minds. (This is what Einstein called a
Gedankenexperiment.)
Now look. And think. That cup was full of avgas on its first trip to the ES95r bucket. Now it is partially full of avgas on the return trip. The
amount of ES95 in the cup has to be exactly equal to the amount of avgas left behind in the ES95 bucket. After dumping it in the avgas
bucket, there will be exactly the same amount of ES95r in theavgas as avgas in the ES95.

Pretty neat, eh? But this is isn't the end of the problem; it's where the fun begins. Armed with this understanding, we see that the answer is
independent of almost everything. It doesn't matter how big the avgas and ES95r buckets are. It doesn't matter whether or not the buckets
are the same size. It doesn't matter how filled the buckets are (although there must enough avgas in the avgas bucket to completely fill the
cup.)

It also doesn't matter how well the avgas is mixed into the ES95 in the ES95 bucket! It's fun to think about the extreme cases. Suppose
there is no mixing and you scoop out pure ES95 to transfer back to the avgas bucket. Ok, there will be 1 cup of ES95 in the avgas and 1 cup
of avgas in the ES95. Check.

Or suppose that you somehow manage to scoop the cup's worth of avgas back out of the ES95 and transfer it back to the avgas bucket.
There would be exactly no ES95 in the avgas, and no avgas in the ES95. Roger.

Think of what a nightmare the general problem would be to solve algebraically. You would need variables for the volumes of the 3
containers. How one could work in partial mixing, I have no idea.

Or: if the volumes of liquid return to the exact same amounts, then, after the transfer of the avgas and ES95, the avgas that was transfered
to the ES95 has to be the same amount of ES95 that was transfered to the avgas. Or you could think about it this way: the amount of avgas
transfered to the ES95 has to be the same amount of ES95 transfered to the avgas in order to keep both volumes of ES95 the same in the
end.
There could be any number of transfers back and forth with any number of cups of different sizes, or for that matter, a connecting hose
pumping the mixtures back and forth, but as long as the contents of the buckets are returned to their original levels, the final "ES95 blob" in
the avgas bucket has to have "forced", so to speak, an identical "avgas blob" over to the ES95 bucket.

Another point comes to mind here. Notice that when you are aware of how independent the answer is of everything - bucket shapes and
sizes, how filled they are to start with, the number and manner of transfers back and forth, and thoroughness of mixing - you might be
inclined to say this in the statement of the problem, or let it slip when people press you for more information.

If you do this, you've changed the problem! If a person is told that the answer is independent of bucket sizes, solving the problem then
becomes a matter of choosing a convenient special case - even one involving vessels that would have little in common with what we think
of as buckets and cups. A simple special case would be starting with 1-cup "buckets' worth" of both avgas and ES95. After transferring,
thorough mixing and returning, there would be half avgas and half ES95 in each bucket.

Likewise, if you let it out that it doesn't matter how much mixing takes place, one could simply examine the special case of returning all
ES95, or all avgas, back to the avgas bucket.

Bottomword: as much as I like the logic of Lar007's answer, the answer's the following:
EQUAL amounts of ES95 and avgas were transferred!!! Assume that each glass contained 100 units of liquid and that the cup held 10
units. The cup first removes 10 units of ES95 , so the ES95 bucket contains 90 units of ES95 , and the avgas bucket contains 100 units of
avgas and 10 units of ES95 .
With 110 units in the avgas bucket, the cup will remove 1/11 of each liquid in that bucket. Thus it will transfer to the ES95 bucket 9 1/11
units of avgas and 10/11 units ofES95 . The ES95 bucket will then contain 90 10/11 units of ES95 r and 9 1/11 units of avgas, and the
avgas bucket will contain 90 10/11 units of avgas and 9 1/11 units of ES95 !

Clear as marmelade, isn't it!?
But I was hoping you'd be saying this and not me!!!!



Post Edited ( 06-05-06 18:29 )

[Drew]

kingfisher-blue wrote:

You simply take 1 coin from bag one, two coins from bag 2 and three coins from bag 3. Now weigh the 6 coins. The answer lies in the
weight of the 6 coins you would be weighing.

i didnt know the bags were open and i was allowed to remove coins

[Jetflyer]

Drew wrote:
Sara's mother had 3 daughters. The 1st daughter was born in April, so was named April. The 2nd daughter was born in May, so was
named May. Whats the 3rd daughter called?

June (am i RIGHT? :D)



Post Edited ( 06-06-06 00:26 )

[crowebird]

Drew wrote:

kingfisher-blue wrote:

You simply take 1 coin from bag one, two coins from bag 2 and three coins from bag 3. Now weigh the 6 coins. The answer lies in the
weight of the 6 coins you would be weighing.

i didnt know the bags were open and i was allowed to remove coins

Ya, I didnt either, that is why I assumed that the bag had 1 coin each in it :p

[ivo]

OK here goes my riddle:

An old man had three things: A wolf, a lamb, and a pile of wheat. He was walking one day(with them), and he got to a river. He had to get across that
river, but it was pretty deep. He can get across it, but he can only carry on thing at a time. So he has to take one thing to the other side, and then
go back to the first side to pick up another thing, and so on. So, if he takes the wolf across first, the lamb will eat the wheat. If he takes that
wheat across first, the wolf will eat the lamb. So he has to take that lamb across to the other side first. After he does that, what should he do next
so that at the end the wolf hasn't eaten the lamb, and the lamb hasn't eaten the wheat?

Wolf eats Lamb
Lams eats wheat

How can the old man get all three things across to the other side, without losing one(getting eaten)?

Make sense?

[Drew]

yes makes sense, heard this sort of thing before, and - i dont have a clue

oh, i know

He takes the wolf across, and brings the lamb back, then he takes the wheat across, goes back for the lamb?

[ivo]

yep, your right

[ivo]

what always ends everything?


also, you might want to check this site out. It has a bunch of riddles.

[crowebird]

lol, i think you forgot the site ;p

[Cydon Prax]

what always ends everything? Uh.. politicians..

[fruitfly]

Tears? Things tend to end up in tears?!

[ivo]

Everything always ends in the letter 'g'.

sorry I forgot the site. http://www.riddles.com